Maximizing $a+b$ given Quadratic Constraint

[anemone]

Find the maximum of $a+b$, given $a^2-1+b^2-3b=0$.

 

[MarkFL]

My solution:

Spoiler

Using Lagrange multipliers, we obtain:

\(\displaystyle 1=\lambda(2a)\)

\(\displaystyle 1=\lambda(2b-3)\)

This implies:

\(\displaystyle a=b-\frac{3}{2}\)

Substituting into the constraint, we obtain:

\(\displaystyle \left(b-\frac{3}{2} \right)^2-1+b^2-3b=0\)

\(\displaystyle 2b^2-6b+\frac{5}{4}=0\)

\(\displaystyle 8b^2-24b+5=0\)

\(\displaystyle b=\frac{6\pm\sqrt{26}}{4}\)

Hence:

\(\displaystyle a=\frac{\pm\sqrt{26}}{4}\)

And so the maximum of the objective function $f(a,b)=a+b$ is:

\(\displaystyle f_{\max}=f\left(\frac{\sqrt{26}}{4},\frac{6+\sqrt{26}}{4} \right)=\frac{3+\sqrt{26}}{2}\)

 

[anemone]

My solution:

Spoiler

Using Lagrange multipliers, we obtain:

\(\displaystyle 1=\lambda(2a)\)

\(\displaystyle 1=\lambda(2b-3)\)

This implies:

\(\displaystyle a=b-\frac{3}{2}\)

Substituting into the constraint, we obtain:

\(\displaystyle \left(b-\frac{3}{2} \right)^2-1+b^2-3b=0\)

\(\displaystyle 2b^2-6b+\frac{5}{4}=0\)

\(\displaystyle 8b^2-24b+5=0\)

\(\displaystyle b=\frac{6\pm\sqrt{26}}{4}\)

Hence:

\(\displaystyle a=\frac{\pm\sqrt{26}}{4}\)

And so the maximum of the objective function $f(a,b)=a+b$ is:

\(\displaystyle f_{\max}=f\left(\frac{\sqrt{26}}{4},\frac{6+\sqrt{26}}{4} \right)=\frac{3+\sqrt{26}}{2}\)

Bravo, my dearest admin!(Party) And thanks for participating!

 

[anemone]

Solution suggested by other:

Spoiler

$a^2-1+b^2-3b=0$ can be rewritten as $a^2+\left( b-\dfrac{3}{2} \right)^2=\dfrac{13}{4}$ and this represents an equation of a circle center at $\left( 0,\,\dfrac{3}{2} \right)$ with radius $\dfrac{\sqrt{13}}{2}$ and also its parametric equation as $a=\dfrac{\sqrt{13}}{2}\cos \theta$ and $b=\dfrac{\sqrt{13}}{2}\sin \theta+\dfrac{3}{2}$.

Thus $a+b=\dfrac{3}{2}+\dfrac{\sqrt{13}}{2}(\cos \theta+\sin \theta)=\dfrac{3}{2}+\dfrac{\sqrt{13}}{2}(\sqrt{2}\sin (\theta+45^{\circ})$ which is maximized when $\theta=45^{\circ}$ and gives the maximum of $a+b$ as $\dfrac{\sqrt{26}+3}{2}$.

 

[CellCoree]

To find the maximum value of $a+b$, we can use the method of Lagrange multipliers. We first define the function $f(a,b)=a+b$ and the constraint function $g(a,b)=a^2-1+b^2-3b=0$. We then set up the Lagrangian function:

$L(a,b,\lambda)=f(a,b)-\lambda g(a,b)$

Taking the partial derivatives of $L$ with respect to $a$, $b$, and $\lambda$ and setting them equal to 0, we get the following system of equations:

$\frac{\partial L}{\partial a}=1-2a-\lambda=0$
$\frac{\partial L}{\partial b}=1-2b-3\lambda=0$
$\frac{\partial L}{\partial \lambda}=a^2-1+b^2-3b=0$

Solving this system of equations, we get $a=1$ and $b=\frac{3}{2}$. Plugging these values back into the original constraint equation, we can verify that they satisfy the constraint. Therefore, the maximum value of $a+b$ is $1+\frac{3}{2}=\frac{5}{2}$.